I'm sure every one of my 34 regular readers (according to Google Analytics!) went immediately to their trusty Texas Instruments TI-35 calculators and used my probabilities from yesterdays' posts to calculate overall odds of making the post season for their favorite teams.
You mean you didn't?
OK, then, let's take a look from the spreadsheet. The beauty of probabilities, statistically speaking is that they can be added, subtracted and multiplied to come up with odds of combinations and odds of mutual exclusivity, among the many other possibilities. Here, we'll really look at the combinatorial property.
Before I delve into this, I want to point out that the best description of what I'm doing comes from
DiamondMind Baseball. I first posted my playoff version of Bill James Log5 theorem in 2005. What I will be doing in the playoffs is exactly what they describe. You can also find my original description
on BigShouldersSports, where I used to occasionally contribute. In essence, I take the home team's winning percentage at home vs. the away teams road winning percentage, all using the pythagorean expectations formula as used during the season, to compute a probability of victory in each game of the series. I then use the combinatorial mathematics to come up with the odds of sweeping, winning two games, or winning just one for each team. As you will see in the playoffs, this also will take into account all five or seven games.
For now, though, we've already had one game, so the odds have changed a little. Looking at the numbers for the Brewers, they have a 49% chance of beating the Cubs twice (since they won last night). The Mets have only a 29% chance of winning two games. The Brewers also have a 33% of winning one game, Mets have a 50% of doing so, facing easier competition. If both teams win one game or both teams win two games, then they end up tied, so we add up both of those probabilities for each team and multiply to get the odds of this happening, which is 64%.
But unfortunately, for the Mets, because they lost yesterday, they do not have the ability to sweep their series and will therefore have to hope that the Brewers will lose at least one game. Otherwise, the chances of the Brewers winning the Wild Card are 59% (odds of sweeping plus odds of winning two games plus odds of winning one game while the Mets lose two plus Mets losing one plus Mets being swept). Those same odds in reverse, i.e. for the Mets to win the WC, are 41%.
If you're wondering why some of those numbers don't total one hundred, it's because these possibilities are mutually exclusive, meaning they exist on their own (this is something I'm sure my regular visitors already know, but I include it in case I get some new visitors that might not). That is not true, though, for the Mets probability (P) of winning, that is exactly 41% because it is calculated 100% (Mets OR Brewers win P) minus 59% (Brewers win P).
Calculating the same for the White Sox and Twins, we get, respectively, 43% and 57%. Best way to check a probability is to do a gut-level check: Twins are playing the Royals, who are the worst team in the league, whereas the Sox are playing the Indians, which are in third place in the division. So the Sox have a tougher schedule.
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